vim search and replace with regex

Question

I'm having trouble getting figuring out a couple of S&Rs for some very large documents I'm editing.

I'd like to replace all instances of [##dname = KOxxx] where xxx is any three-digit number with Variable KOxxx

I came up with [##dname = KO$\d\+$] which correctly finds the text I'd like to replace, but I can't for the life of me figure out how to get the "xxx" to be preserved in the replace. Any ideas?

I'd like to remove the characters > and < around a string. Example: >xxx< where xxx is anywhere from 2-5 characters long.

I had been using the following regex successfully a few months ago, but now it doesn't seem to be working.

%s#>\(\w\{2,5}\)<#\1#g

Any ideas why this might be failing?

Thanks in advance!

\d\+ is one or more numbers, not three numbers. \d{3} would be three numbers. Here \w\{2,5} you are saying the first curly brace is literal, take off the backslash, \w{2,5} to say 2 to 5 word characters (make it a . if you want any character). I'm not sure how vim's find and replace works I'd presume your grouped value for the xxx would be in $1 or \1. — chris85
– chris85, Commented Jun 29, 2015 at 23:02
For future reference, it's better to pose multiple questions separately. — Edward
– Edward, Commented Jun 29, 2015 at 23:21

bjfletcher · Accepted Answer · 2015-06-29 23:12:16Z

1

This is what you're after :)

:%s/\[##dname = KO\(\d*\)\]/Variable KO\1/g

edited Jun 29, 2015 at 23:12

answered Jun 29, 2015 at 23:06

bjfletcher

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2 Comments

chris85 Over a year ago

\d* is not any three-digit number

bjfletcher Over a year ago

yes, s/he can use \d\(3} instead to be more robust

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